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\(=>P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)
CHÚC BẠN HỌC TỐT..........
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}}-\dfrac{5\sqrt{x}+3}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}.\sqrt{x}+3\left(\sqrt{x}+1\right)-\left(5\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+3\sqrt{x}+3-5\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}}-\dfrac{5\sqrt{x}+3}{x+\sqrt{x}}\\ ĐKXĐ:x>0;x\ne1\\ \Rightarrow A=\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{5\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x+3\sqrt{x}+3-5\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Vậy \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với \(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
a,Điều kiện:x\(\ge\)0;x\(\ne\)1
=\(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(\times\)\(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
=\(\dfrac{\sqrt{x}-1_{ }}{\sqrt{x}}\)
b,<=>\(\dfrac{\sqrt{x}_{ }-1}{\sqrt{x}}\)=\(\dfrac{1}{3}\)
<=>3\(\sqrt{x}\)-3=\(\sqrt{x}\)
<=>2\(\sqrt{x}\)=3
<=>x=9/4
a: \(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1-1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}\)
=0
b: \(=\dfrac{3+\sqrt{7}-3+\sqrt{7}}{2}=\dfrac{2\sqrt{7}}{2}=\sqrt{7}\)
c: \(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)
Lời giải:
a)
Áp dụng bất đẳng thức AM-GM:
\(x^3+x^2+x+1\geq 4\sqrt[4]{x^3.x^2.x.1}=4\sqrt[4]{x^6}\)
\(\Rightarrow (x^3+x^2+x+1)^2\geq 16\sqrt{x^6}\)
\(\Leftrightarrow (x^3+x^2+x+1)^2\geq 16x^3\) (đpcm)
Dấu bằng xảy ra khi \(x=1\)
b)
Áp dụng BĐT AM-GM:
\(\frac{b+c}{a}.1\leq \left(\frac{\frac{b+c}{a}+1}{2}\right)^2=\frac{1}{4}\left(\frac{b+c+a}{a}\right)^2\)
\(\Rightarrow \frac{a}{b+c}\geq 4\left(\frac{a}{a+b+c}\right)^2\Leftrightarrow \sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}\)
Thực hiện tương tự với cac phân thức còn lại và cộng theo vế thu được:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq \frac{2a+2b+2c}{a+b+c}=2\)
Dấu bằng xảy ra khi
\(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=1\Rightarrow a+b+c=2a=2b=2c\)
\(\Rightarrow a=b=c\Rightarrow \frac{b+c}{a}=2\neq 1\) (vô lý)
Do đó dấu bằng không xảy ra
Vì vậy: \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\)
1) Đk: \(x\ge4\)
\(\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\sqrt{x-3}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}+x-10}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\sqrt{x^2-16}+x-10=0\)
\(\Leftrightarrow\sqrt{x^2-16}=10-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-16=100-20x+x^2\\x\le10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x=116\\x\le10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\left(N\right)\\x\le10\end{matrix}\right.\)
Kl: x= 29/5
2) Đk: \(x\ge-1\)
\(x^2-5x+14=4\sqrt{x+1}\)
\(\Leftrightarrow x^4+25x^2+196-10x^3-140x+28x^2=16x+16\)
\(\Leftrightarrow x^4-10x^3+53x^2-156x+180=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3-7x^2+32x-60\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x^2-4x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-4x+20=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(N\right)\)
Kl: x=3
a)
Đặt
\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)
Khi đó:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)
\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)
\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)
\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)
Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)
Ta có:
Áp dụng tính chất dãy tỉ số bằng nhau thì:
\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)
Khi đó:
\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)
Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)
Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix}
1\\
-5\end{matrix}\right.\)
a: \(2\sqrt{27}-3\sqrt{54}-\dfrac{1}{3}\sqrt{48}\)
\(=2\cdot3\sqrt{3}-3\cdot3\sqrt{6}-\dfrac{1}{3}\cdot4\sqrt{3}\)
\(=6\sqrt{3}-9\sqrt{6}-\dfrac{4}{3}\sqrt{3}=\dfrac{14}{3}\sqrt[]{3}-9\sqrt{6}\)
b: \(-\dfrac{1}{2}\sqrt{108}+\dfrac{1}{15}\cdot\sqrt{75}-\dfrac{1}{3}\cdot\sqrt{363}\)
\(=-\dfrac{1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{3}\cdot11\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{11}{3}\sqrt{3}=-\dfrac{19}{3}\sqrt{3}\)
c: \(\dfrac{5}{8}\sqrt{48}-\dfrac{1}{33}\cdot\sqrt{363}+\dfrac{3}{14}\cdot\sqrt{147}\)
\(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}=\dfrac{11}{3}\sqrt{3}\)
d:
ĐKXĐ: x>=0; x<>9
Sửa đề:\(\dfrac{x-9}{x-3\sqrt{x}}-\dfrac{x-4}{\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}-\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}+3-x+2\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{-x+3\sqrt{x}+3}{\sqrt{x}}\)
e: ĐKXĐ: x>=0; x<>4
\(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}-\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}\)
\(=\sqrt{x}+1-\sqrt{x}+2=3\)